## Basics of dice probabilities

26 May, 2009 at 7:32 pm (game design, linkedin, mathematics) (, , )

I’ll write a post or a few about probabilities that involve rolling dice. Those who know mathematics might be more interested in probability theory.

I will assume that all probability distributions are discrete and integer-valued. Trying to apply what I say here to continuous distributions will cause problems or require thinking.

# Probability

Probability measures, or indicates, how certain it is that some event will happen (or has happened, in case of imperfect knowledge). Probability of 1 means that something is certain, while 0 means impossibility. Probability 1/2, or $\frac{1}{2}$, or 50%, or 0.5 or in Finnish notation 0,5 means that something happens half the time (if the event is repeated).

I very much prefer working with fractions as they are exact and, in my opinion, more intuitive, but many people like percents. To convert a fraction into percents simply multiply it by hundred and add the %-sign.

An important axiom of probability is that something always happens. The sum of probabilities of all the specific outcomes is 1. By this I mean that if, say, a die is rolled than it gives one and exactly one result. It doesn’t land sideways. It is not hit by a meteor or eaten by a dog.

## Symmetry

Especially when playing around with dice symmetry plays an important role. Symmetric events have the same probability.

I will assume that all dice are fair; in practice they are not and it doesn’t matter. An n-sided die has n symmetric results. All of them hence have the same probability. Something must always happen, so the sum of the probabilities is 1. It follows that for an n-sided die the probability of getting any result from the set $\{ 1 , 2 , \ldots , n-1 , n \}$ is 1/n, while the probability of getting any other integer is zero.

## Notation

Since writing probability all the time gets boring, I’ll use a shorthand: $P( \text{event} ) = p$, which means the probability that event happens is p. For example: $P(\text{d}8=7 ) = 1/8$ and $P(\text{d}8=-4 ) = 0$.

## or, and, not

Some rules for performing calculations with probabilities are in order. First, a definition: Events are independent when knowing something about one of them gives no knowledge about the others. Dice rolls are, as far as this post is concerned, independent: I roll a d12 and get a 1. This tells me nothing about what the next result will be when I roll that d12.

Take two independent events A and B. Now $P(A \text{ and } B ) = P(A)P(B)$. For example: The probability of rolling 1, 2, 3, 4 or 5 (that is: not 6) with a six-sider is 5/6. If we roll two d6s, what is probability of both of them giving a result less than six? Since separate rolls are independent events, this probability is 5/6 times 5/6, which equals 25/36. This rule applies to any finite number of rolls. As long as they are independent, and means multiplication. The independence is not there for show only: Suppose I roll a singe d4. What is the probability of that die giving result of both 1 and 4 at the same time? Obviously, since a given die only gives one result per roll, the event is impossible and hence has probability zero. Careless use of the “and is multiplication”-rule would give 1/4 times 1/4 equals 1/16, which would be wrong.

Multiplying fractions, in case it is not clear: Supposing a, b, c and d are real numbers, b and d are not zero, then $\frac{a}{b}\text{ times } \frac{c}{d} = \frac{ac}{bd}$.

Take any event. Now $P( \text{not event} ) = 1 - P( \text{event} )$. This is a direct consequence of something always happening. Example: The probability of rolling 6 with a d6 is 1/6, from which it follows that the probability of not rolling a 6, which is the probability of rolling something else than 6, is $1 - 1/6 = 5/6$. Now we have the tools for solving one problems with some history: Roll 4d6. Should you bet on rolling at least one 6? The goal here is to determine $P( \text{at least one is 6})$. Using the law of not this problem is the same as determining the probability of none of the dice showing 6, which is same as all of them giving a result from the set $\{ 1 , 2 , 3, 4, 5 \}$. We already know this probability for a single die: It is 5/6. Since separate rolls are made, the events are independent, and hence by the law of and we can simply multiply 5/6 four times, which means raising it to the fourth power: $(5/6)^4 = (5^4)/(6^4) = 625/1296$, which is slightly less than half. By the principle of not we get that the probability of getting at least one 6 is slightly more than half and should be betted on. By symbols the calculation goes as follows: $P(\text{at least one die gives a six}) = 1-P(\text{none of the dice give a six} =$ $1-P(\text{first die is not six and } \dots \text{ and fourth die is not a six}) =$ $1- (P(\text{first die is not a six}) \times \dots \times P(\text{fourth die is not a six}))$ $= 1 - 625/1296 = 1296/1296 - 625/1296 = (1296-625)/ 1296 =$ $671/1296$, which is greater than 648/1296 = 1/2.

Take two events A and B. The probability of at least one of them happening, by which I mean P(A or B), equals the sum of their probabilities minus the probability of A and B both happening; otherwise  the “and” would be counted twice. So, for any events A and B, $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. An important special case: A single d12 is rolled. What is $P(\text{d}12=7 \text{ or } 9 )$?. Since rolling 7 and 9 are clearly distinct events, the probability of both happening with single die roll is 0 (since they never happen at the same time). Hence $P(\text{d}12=7 \text{ or } 9 ) = P(\text{d}12=7) + P(\text{d}12=9) - 0 = 1/6.$ Another useful application: Roll 2d6. What is the probability that at least one of them shows a 6? This can be formulated in another way: What is the probability of first die showing a 6 or the second die showing a 6? Here the events are independent since two dice are cast. Hence, $P(\text{at least one 6}) = P(\text{first d}6=6 \text{ or second d}6=6) =$ $P(\text{first d}6=6) + P(\text{second d}6=6) - P(\text{first and second d}6=6) =$ $1/6 + 1/6 - P(\text{d}6=6)P(\text{d}6=6) = 2/6 - 1/36 = 11/36$.

# More to come?

If someone finds this useful, please say so. I do not know how good I am at expository text like this and I really don’t know the skill level of my audience, if any. A topic I might handle in the future, if anyone is interested, is how to calculate the distribution of a sum of two arbitrary distributions.

I managed to land a quite demanding job, so frequent updates are somewhat unlikely, at least for some time. I’ll need to do some adjusting.